Submission #259608

#	Time	Username	Problem	Language	Result	Execution time	Memory
259608		zaneyu	Arranging Shoes (IOI19_shoes)	C++14	100 / 100	147 ms	20600 KiB

shoes

/*input
2
2 1 -1 -2
*/
//#include "shoes.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<ll,ll>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define MP make_pair
ll mult(ll a,ll b){
    ll res=0LL;
    while(b){
        if(b&1) res=(res+a)%MOD;
        a=(a+a)%MOD;
        b>>=1;
    }
    return res%MOD;
}
ll mypow(ll a,ll b){
    ll res=1LL;
    while(b){
        if(b&1) res=res*a%MOD;
        a=a*a%MOD;
        b>>=1;
    }
    return res;
}
int bit[maxn];
int n;
void upd(int x,int v){
    ++x;
    while(x<=n){
        
        bit[x]+=v;
        x+=lowb(x);
    }
}
int query(int x){
    ++x;
    int ans=0;
    while(x>0){
        ans+=bit[x];
        x-=lowb(x);
    }
    return ans;
}
vector<int> l[maxn],r[maxn];
int pairs[maxn];
bool match[maxn];
long long count_swaps(std::vector<int> s) {
    n=sz(s);
    ll ans=0;
    REP(i,n){
        if(s[i]<0) l[-s[i]].pb(i);
        else r[s[i]].pb(i);
        upd(i,1);
    }
    REP1(i,n/2){
        REP(j,sz(l[i])){
            pairs[l[i][j]]=r[i][j];
            pairs[r[i][j]]=l[i][j];
            if(l[i][j]>r[i][j]) ++ans;
        }
    }
    REP(i,n){
        if(match[i]) continue;
        match[i]=match[pairs[i]]=1;
        upd(i,-1);
        upd(pairs[i],-1);
        ans+=query(pairs[i]);
    }
    return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 15 / 15